Question

Will award medal
Question posted below, Center of mass

Answer 1

@SolomonZelman

Answer 2

@SithsAndGiggles

Answer 3

\[\int\limits_{0}^{1}\int\limits_{0}^{x}3x^2dydx\] this was my double integral for the mass

Answer 4

the double integral for the mass is
\[\int\limits_{x = 0}^{2} \ \ \int\limits_{y = x}^{-2x + 6} 3 x^2 dy\ dx = \int\limits_{x = 0}^{2} \ 3 x^2 [(6-2x) - x] dx\]

and that does give 12

the moment integral, calculating in the x direction:

\[M_x = \int\limits_{x =0}^{2}\int\limits_{y = x}^{-2x + 6} 3 x^3 dx dy = \int\limits_{x =0}^{2}\int\limits_{y = x}^{-2x + 6} 3 x^3 dx dy \]

gives 72/5 in the x direction. which suggest 1.2 for COMx

you would have to split the integral in the other direction to get COMy.

Answer 5

\[M_y = \int\limits_{x = 0}^{2}\int\limits_{y = x}^{-2x +6} 3x^2 y \ dx\ dy = \int\limits_{0}^{2} \left| \frac{3x^2y^2}{2} \right|^{-2x+6}_{x} dx\]
\[\int\limits_{0}^{2} \frac{9}{2}x^4 - 36 x^3 + 54 x^2 \ dx = \left | \frac{9}{10}x^5 - 9 x^4 + 18x^3 \right |^{2}_{0} = \frac{192}{5}\]

COMy = 192/(5*12) = 12/5 = 2.4